Khalil Jones

2023-03-14

The resistance of a wire of iron is 10 ohms and temp. coefficient of resistivity is 5 \times 10^{−3}/^{\circ} C. At 20^{\circ} C it carries 30 milliamperes of current. Keeping constant potential difference between its ends, the temperature of the wire is raised to 120^{\circ} C. The current in milliamperes that flows in the wire is

Veronella6nk

Beginner2023-03-15Added 6 answers

Step 1: Calculation of resistance at ${120}^{\circ}\text{}C$

Let variables at ${20}^{\circ}\text{}C$ be marked by subscript 1 and at ${120}^{\circ}\text{}C$ be marked by subscript 2.

Given that,

$\frac{{R}_{1}}{{R}_{2}}=\frac{1+\alpha {t}_{1}}{1+\alpha {t}_{2}}$

Where ${t}_{1}={20}^{\circ}C$ and ${R}_{1}$ is resistance at ${t}_{1}$ and ${t}_{2}={120}^{\circ}C$ and ${R}_{2}$ is resistance at ${t}_{2}$.

We have,

${R}_{1}=10\mathrm{\Omega},\alpha =5\times {10}^{-3}{/}^{\circ}C\phantom{\rule{0ex}{0ex}}\therefore \frac{10}{{R}_{2}}=\frac{1+(5\times {10}^{-3}\times 20)}{1+(5\times {10}^{-3}\times 120)}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{2}\approx 15\mathrm{\Omega}$

Step 2: Calculation of current at ${120}^{\circ}C$

when the potential difference is constant,

$i\propto \frac{1}{R}\phantom{\rule{0ex}{0ex}}\therefore \frac{{i}_{1}}{{i}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{15}{10}=1.5$

It is also provided,

$i1=30mA\phantom{\rule{0ex}{0ex}}\therefore i2=301.5\phantom{\rule{0ex}{0ex}}\Rightarrow i2=20mA$

Therefore, current flowing through the circuit at ${120}^{\circ}$ C is 20mA.

Let variables at ${20}^{\circ}\text{}C$ be marked by subscript 1 and at ${120}^{\circ}\text{}C$ be marked by subscript 2.

Given that,

$\frac{{R}_{1}}{{R}_{2}}=\frac{1+\alpha {t}_{1}}{1+\alpha {t}_{2}}$

Where ${t}_{1}={20}^{\circ}C$ and ${R}_{1}$ is resistance at ${t}_{1}$ and ${t}_{2}={120}^{\circ}C$ and ${R}_{2}$ is resistance at ${t}_{2}$.

We have,

${R}_{1}=10\mathrm{\Omega},\alpha =5\times {10}^{-3}{/}^{\circ}C\phantom{\rule{0ex}{0ex}}\therefore \frac{10}{{R}_{2}}=\frac{1+(5\times {10}^{-3}\times 20)}{1+(5\times {10}^{-3}\times 120)}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{2}\approx 15\mathrm{\Omega}$

Step 2: Calculation of current at ${120}^{\circ}C$

when the potential difference is constant,

$i\propto \frac{1}{R}\phantom{\rule{0ex}{0ex}}\therefore \frac{{i}_{1}}{{i}_{2}}=\frac{{R}_{2}}{{R}_{1}}=\frac{15}{10}=1.5$

It is also provided,

$i1=30mA\phantom{\rule{0ex}{0ex}}\therefore i2=301.5\phantom{\rule{0ex}{0ex}}\Rightarrow i2=20mA$

Therefore, current flowing through the circuit at ${120}^{\circ}$ C is 20mA.